Two point charges +3 C and +6 C separated by a distance 'r' , repel...

Initial charges are $q_1 = +3\text{ C}$ and $q_2 = +6\text{ C}$.

Initial force $F_1 = \dfrac{k q_1 q_2}{r^2} = \dfrac{k (3)(6)}{r^2} = \dfrac{18k}{r^2} = 27\text{ N}$.

This gives $\dfrac{k}{r^2} = \dfrac{27}{18} = 1.5\text{ N}$.

When an additional charge of $-5\text{ C}$ is given to each, the new charges become:

$q_1' = +3 - 5 = -2\text{ C}$

$q_2' = +6 - 5 = +1\text{ C}$

The new force $F_2 = \dfrac{k |q_1'| |q_2'|}{r^2} = \dfrac{k (2)(1)}{r^2} = \dfrac{2k}{r^2}$.

Substituting the value of $\dfrac{k}{r^2}$, we get:

$F_2 = 2 \times 1.5 = 3\text{ N}$.

Since the new charges have opposite signs, the force between them is attractive.

Answer: $3\text{N}$ (attractive)